Inequalities in the Lowner Partial Order

1. Inequalities in the L¨ owner Partial Order

Throughout we consider square complex matrices. Since rectangular matrices can be augmented to square ones with zero blocks, all the results on singular values and unitarily invariant norms hold as well for rectangular matrices. Denote by Mn the space of n×n complex matrices. A matrix A ∈ Mn is often regarded
as a linear operator on Cn endowed with the usual inner product x, y ≡ j xj y¯j for x = (xj ), y = (yj ) ∈ Cn . Then the conjugate transpose A∗ is the adjoint of A. The Euclidean norm on Cn is x = x, x1/2 . A matrix A ∈ Mn is called positive semidefinite if Ax, x ≥ 0

for all x ∈ Cn .

(1.1)

Thus for a positive semidefinite A, Ax, x = x, Ax. For any A ∈ Mn and x, y ∈ Cn , we have 4Ax, y =

3 

ik A(x + ik y), x + ik y,

k=0

4x, Ay =

3 

ik x + ik y, A(x + ik y)

k=0

√

where i = −1. It is clear from these two identities that the condition (1.1) implies A∗ = A. Therefore a positive semidefinite matrix is necessarily Hermitian. In the sequel when we talk about matrices A, B, C, . . . without specifying their orders, we always mean that they are of the same order. For Hermitian matrices G, H we write G ≤ H or H ≥ G to mean that H − G is positive semidefinite. In particular, H ≥ 0 indicates that H is positive semidefinite. This is known as the L¨ owner partial order; it is induced in the real space of (complex) Hermitian matrices by the cone of positive semidefinite matrices. If H is positive definite, that is, positive semidefinite and invertible, we write H > 0. Let f (t) be a continuous real-valued function defined on a real interval Ω and H be a Hermitian matrix with eigenvalues in Ω. Let H = U diag(λ1 , . . . , λn )U ∗ be a spectral decomposition with U unitary. Then the functional calculus for H is defined as

X. Zhan: LNM 1790, pp. 1–15, 2002. c Springer-Verlag Berlin Heidelberg 2002


2

1. The L¨ owner Partial Order

f (H) ≡ U diag(f (λ1 ), . . . , f (λn ))U ∗ .

(1.2)

This is well-defined, that is, f (H) does not depend on particular spectral decompositions of H. To see this, first note that (1.2) coincides with the usual
k
k polynomial calculus: If f (t) = j=0 cj tj then f (H) = j=0 cj H j . Second, by the Weierstrass approximation theorem, every continuous function on a finite closed interval Ω is uniformly approximated by a sequence of polynomials. Here we need the notion of a norm on matrices to give a precise meaning of approximation by a sequence of matrices. We denote by A∞ the spectral (operator) norm of A: A∞ ≡ max{Ax : x = 1, x ∈ Cn }. The spectral norm is submultiplicative: AB∞ ≤ A∞ B∞ . The positive semidefinite square root H 1/2 of H ≥ 0 plays an important role. Some results in this chapter are the basis of inequalities for eigenvalues, singular values and norms developed in subsequent chapters. We always use capital letters for matrices and small letters for numbers unless otherwise stated.

1.1 The L¨ owner-Heinz inequality Denote by I the identity matrix. A matrix C is called a contraction if C ∗ C ≤ I, or equivalently, C∞ ≤ 1. Let ρ(A) be the spectral radius of A. Then ρ(A) ≤ A∞ . Since AB and BA have the same eigenvalues, ρ(AB) = ρ(BA). Theorem 1.1 (L¨ owner-Heinz) If A ≥ B ≥ 0 and 0 ≤ r ≤ 1 then Ar ≥ B r .

(1.3)

Proof. The standard continuity argument is that in many cases, e.g., the present situation, to prove some conclusion on positive semidefinite matrices it suffices to show it for positive definite matrices by considering A + I,  ↓ 0. Now we assume A > 0. Let ∆ be the set of those r ∈ [0, 1] such that (1.3) holds. Obviously 0, 1 ∈ ∆ and ∆ is closed. Next we show that ∆ is convex, from which follows ∆ = [0, 1] and the proof will be completed. Suppose s, t ∈ ∆. Then A−s/2 B s A−s/2 ≤ I,

A−t/2 B t A−t/2 ≤ I

or equivalently B s/2 A−s/2 ∞ ≤ 1, B t/2 A−t/2 ∞ ≤ 1. Therefore A−(s+t)/4 B (s+t)/2 A−(s+t)/4 ∞ = ρ(A−(s+t)/4 B (s+t)/2 A−(s+t)/4 ) = ρ(A−s/2 B (s+t)/2 A−t/2 ) = A−s/2 B (s+t)/2 A−t/2 ∞ = (B s/2 A−s/2 )∗ (B t/2 A−t/2 )∞ ≤ B s/2 A−s/2 ∞ B t/2 A−t/2 ∞ ≤ 1.

1.1 The L¨ owner-Heinz inequality

3

Thus A−(s+t)/4 B (s+t)/2 A−(s+t)/4 ≤ I and consequently B (s+t)/2 ≤ A(s+t)/2 , i.e., (s + t)/2 ∈ ∆. This proves the convexity of ∆. 

How about this theorem for r > 1? The answer is negative in general. The example       43 21 10 2 2 A= , B= , A −B = 32 11 00 shows that A ≥ B ≥ 0 ⇒ A2 ≥ B 2 . The next result gives a conceptual understanding, and this seems a typical way of mathematical thinking. We will have another occasion in Section 4.6 to mention the notion of a C ∗ algebra, but for our purpose it is just Mn . Let A be a Banach space over C. If A is also an algebra in which the norm is submultiplicative: AB ≤ A B, then A is called a Banach algebra. An involution on A is a map A → A∗ of A into itself such that for all A, B ∈ A and α ∈ C (i) (A∗ )∗ = A;

(ii) (AB)∗ = B ∗ A∗ ;

(iii) (αA + B)∗ = α ¯ A∗ + B ∗ .

A C ∗ -algebra A is a Banach algebra with involution such that A∗ A = A2

for all A ∈ A.

An element A ∈ A is called positive if A = B ∗ B for some B ∈ A. It is clear that Mn with the spectral norm and with conjugate transpose being the involution is a C ∗ -algebra. Note that the L¨ owner-Heinz inequality also holds for elements in a C ∗ -algebra and the same proof works, since every fact used there remains true, for instance, ρ(AB) = ρ(BA). Every element T ∈ A can be written uniquely as T = A + iB with A, B Hermitian. In fact A = (T + T ∗ )/2, B = (T − T ∗ )/2i. This is called the Cartesian decomposition of T. We say that A is commutative if AB = BA for all A, B ∈ A. Theorem 1.2 Let A be a C ∗ -algebra and r > 1. If A ≥ B ≥ 0, A, B ∈ A implies Ar ≥ B r , then A is commutative. Proof. Since r > 1, there exists a positive integer k such that rk > 2. Suppose k k A ≥ B ≥ 0. Use the assumption successively k times we get Ar ≥ B r . k Then apply the L¨ owner-Heinz inequality with the power 2/r < 1 to obtain A2 ≥ B 2 . Therefore it suffices to prove the theorem for the case r = 2. For any A, B ≥ 0 and  > 0 we have A + B ≥ A. Hence by assumption, (A + B)2 ≥ A2 . This yields AB + BA + B 2 ≥ 0 for any  > 0. Thus AB + BA ≥ 0

for all A, B ≥ 0.

(1.4)

Let AB = G + iH with G, H Hermitian. Then (1.4) means G ≥ 0. Applying this to A, BAB,

4

1. The L¨ owner Partial Order

A(BAB) = G2 − H 2 + i(GH + HG)

(1.5)

gives G2 ≥ H 2 . So the set Γ ≡ {α ≥ 1 : G2 ≥ αH 2 for all A, B ≥ 0 with AB = G + iH} where G + iH is the Cartesian decomposition, is nonempty. Suppose Γ is bounded. Then since Γ is closed, it has a largest element λ. By (1.4) H 2 (G2 − λH 2 ) + (G2 − λH 2 )H 2 ≥ 0, i.e., G2 H 2 + H 2 G2 ≥ 2λH 4 .

(1.6)

From (1.5) we have (G2 − H 2 )2 ≥ λ(GH + HG)2 , i.e., G4 + H 4 − (G2 H 2 + H 2 G2 ) ≥ λ[GH 2 G + HG2 H + G(HGH) + (HGH)G]. Combining this inequality, (1.6) and the inequalities GH 2 G ≥ 0, G(HGH) + (HGH)G ≥ 0 (by (1.4) and G ≥ 0), HG2 H ≥ λH 4 (by the definition of λ) we obtain G4 ≥ (λ2 + 2λ − 1)H 4 . Then applying the L¨ owner-Heinz inequality again we get G2 ≥ (λ2 + 2λ − 1)1/2 H 2 for all G, H in the Cartesian decomposition AB = G + iH with A, B ≥ 0. Hence (λ2 + 2λ − 1)1/2 ∈ Γ , which yields (λ2 + 2λ − 1)1/2 ≤ λ by definition. Consequently λ ≤ 1/2. This contradicts the assumption that λ ≥ 1. So Γ is unbounded and G2 ≥ αH 2 for all α ≥ 1, which is possible only when H = 0. Consequently AB = BA for all positive A, B. Finally by the Cartesian decomposition and the fact that every Hermitian element is a difference of two positive elements we conclude that XY = Y X for all X, Y ∈ A. 

Since Mn is noncommutative when n ≥ 2, we know that for any r > 1 there exist A ≥ B ≥ 0 but Ar ≥ B r . Notes and References. The proof of Theorem 1.1 here is given by G. K. Pedersen [79]. Theorem 1.2 is due to T. Ogasawara [77].

1.2 Maps on Matrix Spaces A real-valued continuous function f (t) defined on a real interval Ω is said to be operator monotone if A≤B

implies

f (A) ≤ f (B)

1.2 Maps on Matrix Spaces

5

for all such Hermitian matrices A, B of all orders whose eigenvalues are contained in Ω. f is called operator convex if for any 0 < λ < 1, f (λA + (1 − λ)B) ≤ λf (A) + (1 − λ)f (B) holds for all Hermitian matrices A, B of all orders with eigenvalues in Ω. f is called operator concave if −f is operator convex. Thus the L¨ owner-Heinz inequality says that the function f (t) = tr , (0 < r ≤ 1) is operator monotone on [0, ∞). Another example of operator monotone function is log t on (0, ∞) while an example of operator convex function is g(t) = tr on (0, ∞) for −1 ≤ r ≤ 0 or 1 ≤ r ≤ 2 [17, p.147]. If we know the formula  sin rπ ∞ sr−1 t r ds (0 < r < 1) t = π s+t 0 then Theorem 1.1 becomes quite obvious. In general we have the following useful integral representations for operator monotone and operator convex functions. This is part of L¨ owner’s deep theory [17, p.144 and 147] (see also [32]). Theorem 1.3 If f is an operator monotone function on [0, ∞), then there exists a positive measure µ on [0, ∞) such that  ∞ st f (t) = α + βt + dµ(s) (1.7) s +t 0 where α is a real number and β ≥ 0. If g is an operator convex function on [0, ∞) then there exists a positive measure µ on [0, ∞) such that  ∞ st2 2 dµ(s) (1.8) g(t) = α + βt + γt + s+t 0 where α, β are real numbers and γ ≥ 0. The three concepts of operator monotone, operator convex and operator concave functions are intimately related. For example, a nonnegative continuous function on [0, ∞) is operator monotone if and only if it is operator concave [17, Theorem V.2.5]. A map Φ : Mm → Mn is called positive if it maps positive semidefinite matrices to positive semidefinite matrices: A ≥ 0 ⇒ Φ(A) ≥ 0. Denote by In the identity matrix in Mn . Φ is called unital if Φ(Im ) = In . We will first derive some inequalities involving unital positive linear maps, operator monotone functions and operator convex functions, then use these results to obtain inequalities for matrix Hadamard products. The following fact is very useful. Lemma 1.4 Let A > 0. Then

6

1. The L¨ owner Partial Order



A B B∗ C

 ≥0

if and only if the Schur complement C − B ∗ A−1 B ≥ 0. Lemma 1.5 Let Φ be a unital positive linear map from Mm to Mn . Then Φ(A2 ) ≥ Φ(A)2 Φ(A−1 ) ≥ Φ(A)−1

(A ≥ 0), (A > 0).

(1.9) (1.10)


m Proof. Let A = j=1 λj Ej be the spectral decomposition of A, where λj ≥ 0 (j = 1, . . . , m) are the eigenvalues and
Ej (j = 1, . . . , m) are the m corresponding eigenprojections of rank one with j=1 Ej = Im . Then since

m m A2 = j=1 λ2j Ej and by unitality In = Φ(Im ) = j=1 Φ(Ej ), we have 

   m  1 λj In Φ(A) ⊗ Φ(Ej ), = λj λ2j Φ(A) Φ(A2 ) j=1

where ⊗ denotes the Kronecker (tensor) product. Since   1 λj ≥0 λj λ2j and by positivity Φ(Ej ) ≥ 0 (j = 1, . . . , m), we have   1 λj ⊗ Φ(Ej ) ≥ 0, λj λ2j j = 1, . . . , m. Consequently 

 In Φ(A) ≥0 Φ(A) Φ(A2 )

which implies (1.9) by Lemma 1.4. In a similar way, using   λj 1 ≥0 1 λ−1 j we can conclude that



 Φ(A) In ≥0 In Φ(A−1 )

which implies (1.10) again by Lemma 1.4.



Theorem 1.6 Let Φ be a unital positive linear map from Mm to Mn and f an operator monotone function on [0, ∞). Then for every A ≥ 0,

1.2 Maps on Matrix Spaces

7

f (Φ(A)) ≥ Φ(f (A)). Proof. By the integral representation (1.7) it suffices to prove Φ(A)[sI + Φ(A)]−1 ≥ Φ[A(sI + A)−1 ],

s > 0.

Since A(sI + A)−1 = I − s(sI + A)−1 and similarly for the left side, this is equivalent to [Φ(sI + A)]−1 ≤ Φ[(sI + A)−1 ] 

which follows from (1.10).

Theorem 1.7 Let Φ be a unital positive linear map from Mm to Mn and g an operator convex function on [0, ∞). Then for every A ≥ 0, g(Φ(A)) ≤ Φ(g(A)). Proof. By the integral representation (1.8) it suffices to show Φ(A)2 ≤ Φ(A2 ) and

(1.11)

Φ(A)2 [sI + Φ(A)]−1 ≤ Φ[A2 (sI + A)−1 ],

s > 0.

(1.12)

(1.11) is just (1.9). Since A2 (sI + A)−1 = A − sI + s2 (sI + A)−1 , Φ(A)2 [sI + Φ(A)]−1 = Φ(A) − sI + s2 [sI + Φ(A)]−1 , (1.12) follows from (1.10). This completes the proof.



Since f1 (t) = tr (0 < r ≤ 1) and f2 (t) = log t are operator monotone functions on [0, ∞) and (0, ∞) respectively, g(t) = tr is operator convex on (0, ∞) for −1 ≤ r ≤ 0 and 1 ≤ r ≤ 2, from Theorems 1.6, 1.7 we get the following corollary. Corollary 1.8 Let Φ be a unital positive linear map from Mm to Mn . Then Φ(Ar ) ≤ Φ(A)r , Φ(Ar ) ≥ Φ(A)r ,

A ≥ 0,

0 < r ≤ 1;

A > 0, −1 ≤ r ≤ 0 or 1 ≤ r ≤ 2;

Φ(log A) ≤ log(Φ(A)),

A > 0.

Given A = (aij ), B = (bij ) ∈ Mn , the Hadamard product of A and B is defined as the entry-wise product: A ◦ B ≡ (aij bij ) ∈ Mn . For this topic see

8

1. The L¨ owner Partial Order

[52, Chapter 5]. We denote by A[α] the principal submatrix of A indexed by α. The following simple observation is very useful. Lemma 1.9 For any A, B ∈ Mn , A ◦ B = (A ⊗ B)[α] where α = {1, n + 2, 2n + 3, . . . , n2 }. Consequently there is a unital positive linear map Φ from Mn2 to Mn such that Φ(A ⊗ B) = A ◦ B for all A, B ∈ Mn . As an illustration of the usefulness of this lemma, consider the following reasoning: If A, B ≥ 0, then evidently A ⊗ B ≥ 0. Since A ◦ B is a principal submatrix of A ⊗ B, A ◦ B ≥ 0. Similarly A ◦ B > 0 for the case when both A and B are positive definite. In other words, the Hadamard product of positive semidefinite (definite) matrices is positive semidefinite (definite). This important fact is known as the Schur product theorem. Corollary 1.10 Ar ◦ B r ≤ (A ◦ B)r ,

A, B ≥ 0,

0 < r ≤ 1;

Ar ◦ B r ≥ (A ◦ B)r , A, B > 0, −1 ≤ r ≤ 0 or 1 ≤ r ≤ 2; (log A + log B) ◦ I ≤ log(A ◦ B),

A, B > 0.

(1.13) (1.14) (1.15)

Proof. This is an application of Corollary 1.8 with A there replaced by A⊗B and Φ being defined in Lemma 1.9. For (1.13) and (1.14) just use the fact that (A ⊗ B)t = At ⊗ B t for real number t. See [52] for properties of the Kronecker product. For (1.15) we have d d (A ⊗ B)t |t=0 = (At ⊗ B t )|t=0 dt dt = (log A) ⊗ I + I ⊗ (log B).

log(A ⊗ B) =

This can also be seen by using the spectral decompositions of A and B.



We remark that the inequality in (1.14) is also valid for A, B ≥ 0 in the case 1 ≤ r ≤ 2. Given a positive integer k, let us denote the kth Hadamard power of A = (aij ) ∈ Mn by A(k) ≡ (akij ) ∈ Mn . Here are two interesting consequences of Corollary 1.10: For every positive integer k, (Ar )(k) ≤ (A(k) )r , (Ar )(k) ≥ (A(k) )r ,

A ≥ 0,

0 < r ≤ 1;

A > 0, −1 ≤ r ≤ 0 or 1 ≤ r ≤ 2.

Corollary 1.11 For A, B ≥ 0, the function f (t) = (At ◦ B t )1/t is increasing on [1, ∞), i.e., (As ◦ B s )1/s ≤ (At ◦ B t )1/t ,

1 ≤ s < t.

1.2 Maps on Matrix Spaces

9

Proof. By Corollary 1.10 we have As ◦ B s ≤ (At ◦ B t )s/t . Then applying the L¨ owner-Heinz inequality with the power 1/s yields the conclusion. 

Let Pn be the set of positive semidefinite matrices in Mn . A map Ψ from Pn × Pn into Pm is called jointly concave if Ψ (λA + (1 − λ)B, λC + (1 − λ)D) ≥ λΨ (A, C) + (1 − λ)Ψ (B, D) for all A, B, C, D ≥ 0 and 0 < λ < 1. For A, B > 0, the parallel sum of A and B is defined as A : B = (A−1 + B −1 )−1 . Note that A : B = A − A(A + B)−1 A and 2(A : B) = {(A−1 + B −1 )/2}−1 is the harmonic mean of A, B. Since A : B decreases as A, B decrease, we can define the parallel sum for general A, B ≥ 0 by A : B = lim{(A + I)−1 + (B + I)−1 }−1 .
↓0

Using Lemma 1.4 it is easy to verify that     A+B A A : B = max X ≥ 0 : ≥0 A A−X where the maximum is with respect to the L¨owner partial order. From this extremal representation it follows readily that the map (A, B) → A : B is jointly concave. Lemma 1.12 For 0 < r < 1 the map (A, B) → Ar ◦ B 1−r is jointly concave in A, B ≥ 0. Proof. It suffices to prove that the map (A, B) → Ar ⊗ B 1−r is jointly concave in A, B ≥ 0, since then the assertion will follow via Lemma 1.9. We may assume B > 0. Using Ar ⊗ B 1−r = (A ⊗ B −1 )r (I ⊗ B) and the integral representation  sin rπ ∞ sr−1 t r ds (0 < r < 1) t = π s+t 0 we get

10

1. The L¨ owner Partial Order

Ar ⊗ B 1−r =

sin rπ π

 0

∞

sr−1 (A ⊗ B −1 )(A ⊗ B −1 + sI ⊗ I)−1 (I ⊗ B)ds.

Since A ⊗ B −1 and I ⊗ B commute, it is easy to see that (A ⊗ B −1 )(A ⊗ B −1 + sI ⊗ I)−1 (I ⊗ B) = (s−1 A ⊗ I) : (I ⊗ B). We know that the parallel sum is jointly concave. Thus the integrand above is also jointly concave, and so is Ar ⊗ B 1−r . This completes the proof. 

Corollary 1.13 For A, B, C, D ≥ 0 and p, q > 1 with 1/p + 1/q = 1, A ◦ B + C ◦ D ≤ (Ap + C p )1/p ◦ (B q + Dq )1/q . Proof. This is just the mid-point joint concavity case λ = 1/2 of Lemma 1.12 with r = 1/p. 

Let f (x) be a real-valued differentiable function defined on some real interval. We denote by ∆f (x, y) ≡ [f (x)−f (y)]/(x−y) the difference quotient where ∆f (x, x) ≡ f  (x). Let H(t) ∈ Mn be a family of Hermitian matrices for t in an open real interval (a, b) and suppose the eigenvalues of H(t) are contained in some open real interval Ω for all t ∈ (a, b). Let H(t) = U (t)Λ(t)U (t)∗ be the spectral decomposition with U (t) unitary and Λ(t) = diag(λ1 (t), . . . , λn (t)). Assume that H(t) is continuously differentiable on (a, b) and f : Ω → R is a continuously differentiable function. Then it is known [52, Theorem 6.6.30] that f (H(t)) is continuously differentiable and d f (H(t)) = U (t){[∆f (λi (t), λj (t))] ◦ [U (t)∗ H  (t)U (t)]}U (t)∗ . dt Theorem 1.14 For A, B ≥ 0 and p, q > 1 with 1/p + 1/q = 1, A ◦ B ≤ (Ap ◦ I)1/p (B q ◦ I)1/q . Proof. Denote

C ≡ (Ap ◦ I)1/p ≡ diag(λ1 , . . . , λn ), D ≡ (B q ◦ I)1/q ≡ diag(µ1 , . . . , µn ).

By continuity we may assume that λi = λj and µi = µj for i = j. Using the above differential formula we compute   d p (C + tAp )1/p  = X ◦ Ap dt t=0

1.3 Inequalities for Matrix Powers

11

  d q q 1/q  (D + tB )  = Y ◦ Bq dt t=0

and

where X = (xij ) and Y = (yij ) are defined by , xij = (λi − λj )(λpi − λpj )−1 for i = j and xii = p−1 λ1−p i yij = (µi − µj )(µqi − µqj )−1 for i = j and yii = q −1 µ1−q . i By Corollary 1.13 C ◦ D + tA ◦ B ≤ (C p + tAp )1/p ◦ (Dq + tB q )1/q for any t ≥ 0. Therefore, via differentiation at t = 0 we have d p (C + tAp )1/p ◦ (Dq + tB q )1/q |t=0 dt = X ◦ Ap ◦ D + C ◦ Y ◦ B q

A◦B ≤

= (X ◦ I)(Ap ◦ I)D + C(Y ◦ I)(B q ◦ I) = p−1 C 1−p (Ap ◦ I)D + q −1 CD1−q (B q ◦ I) = (Ap ◦ I)1/p (B q ◦ I)1/q . 

This completes the proof.

We will need the following result in the next section and in Chapter 3. See [17] for a proof. Theorem 1.15 Let f be an operator monotone function on [0, ∞), g an operator convex function on [0, ∞) with g(0) ≤ 0. Then for every contraction C, i.e., C∞ ≤ 1 and every A ≥ 0, f (C ∗ AC) ≥ C ∗ f (A)C,

(1.16)

g(C ∗ AC) ≤ C ∗ g(A)C.

(1.17)

Notes and References. As already remarked, Theorem 1.3 is part of the L¨owner theory. The inequality (1.16) in Theorem 1.15 is due to F. Hansen [43] while the inequality (1.17) is proved by F. Hansen and G. K. Pedersen [44]. All other results in this section are due to T. Ando [3, 8].

1.3 Inequalities for Matrix Powers The purpose of this section is to prove the following result.

12

1. The L¨ owner Partial Order

Theorem 1.16 If A ≥ B ≥ 0 then

and

(B r Ap B r )1/q ≥ B (p+2r)/q

(1.18)

A(p+2r)/q ≥ (Ar B p Ar )1/q

(1.19)

for r ≥ 0, p ≥ 0, q ≥ 1 with (1 + 2r)q ≥ p + 2r. Proof. We abbreviate “the L¨ owner-Heinz inequality” to LH, and first prove (1.18). If 0 ≤ p < 1, then by LH, Ap ≥ B p and hence B r Ap B r ≥ B p+2r . Applying LH again with the power 1/q gives (1.18). Next we consider the case p ≥ 1. It suffices to prove (B r Ap B r )(1+2r)/(p+2r) ≥ B 1+2r for r ≥ 0, p ≥ 1, since by assumption q ≥ (p + 2r)/(1 + 2r), and then (1.18) follows from this inequality via LH. Let us introduce t to write the above inequality as 1 + 2r . (1.20) (B r Ap B r )t ≥ B 1+2r , t = p + 2r Note that 0 < t ≤ 1, as p ≥ 1. We will show (1.20) by induction on k = 0, 1, 2, . . . for the intervals (2k−1 − 1/2, 2k − 1/2] containing r. Since (0, ∞) = k−1 ∪∞ − 1/2, 2k − 1/2], (1.20) is proved. k=0 (2 By the standard continuity argument, we may and do assume that A, B are positive definite. First consider the case k = 0, i.e., 0 < r ≤ 1/2. By LH A2r ≥ B 2r and hence B r A−2r B r ≤ I, which means that A−r B r is a contraction. Applying (1.16) in Theorem 1.15 with f (x) = xt yields (B r Ap B r )t = [(A−r B r )∗ Ap+2r (A−r B r )]t ≥ (A−r B r )∗ A(p+2r)t (A−r B r ) = B r AB r ≥ B 1+2r , proving (1.20) for the case k = 0. Now suppose that (1.20) is true for r ∈ (2k−1 − 1/2, 2k − 1/2]. Denote A1 = (B r Ap B r )t , B1 = B 1+2r . Then our assumption is A1 ≥ B1

with t =

1 + 2r . p + 2r

Since p1 ≡ 1/t ≥ 1, apply the already proved case r1 ≡ 1/2 to A1 ≥ B1 to get 1 + 2r1 (B1r1 Ap11 B1r1 )t1 ≥ B11+2r1 , t1 ≡ . (1.21) p1 + 2r1 2+4r . Denote s = 2r + 1/2. We have s ∈ (2k − 1/2, 2k+1 − Note that t1 = p+4r+1 1/2]. Then explicitly (1.21) is

1.4 Block Matrix Techniques

(B s Ap B s )t1 ≥ B 1+2s ,

t1 =

13

1 + 2s , p + 2s

which shows that (1.20) holds for r ∈ (2k − 1/2, 2k+1 − 1/2]. This completes the inductive argument and (1.18) is proved. A ≥ B > 0 implies B −1 ≥ A−1 > 0. In (1.18) replacing A, B by B −1 , A−1 respectively yields (1.19). 

The case q = p ≥ 1 of Theorem 1.16 is the following Corollary 1.17 If A ≥ B ≥ 0 then (B r Ap B r )1/p ≥ B (p+2r)/p , A(p+2r)/p ≥ (Ar B p Ar )1/p for all r ≥ 0 and p ≥ 1. A still more special case is the next Corollary 1.18 If A ≥ B ≥ 0 then (BA2 B)1/2 ≥ B 2

and

A2 ≥ (AB 2 A)1/2 .

At first glance, Corollary 1.18 (and hence Theorem 1.16) is strange: For positive numbers a ≥ b, we have a2 ≥ (ba2 b)1/2 ≥ b2 . We know the matrix analog that A ≥ B ≥ 0 implies A2 ≥ B 2 is false, but Corollary 1.18 asserts that the matrix analog of the stronger inequality (ba2 b)1/2 ≥ b2 holds. This example shows that when we move from the commutative world to the noncommutative one, direct generalizations may be false, but a judicious modification may be true. Notes and References. Corollary 1.18 is a conjecture of N. N. Chan and M. K. Kwong [29]. T. Furuta [38] solved this conjecture by proving the more general Theorem 1.16. See [39] for a related result.

1.4 Block Matrix Techniques In the proof of Lemma 1.5 we have seen that block matrix arguments are powerful. Here we give one more example. In later chapters we will employ other types of block matrix techniques. Theorem 1.19 Let A, B, X, Y be matrices with A, B positive definite and X, Y arbitrary. Then (X ∗ A−1 X) ◦ (Y ∗ B −1 Y ) ≥ (X ◦ Y )∗ (A ◦ B)−1 (X ◦ Y ),

(1.22)

14

1. The L¨ owner Partial Order

X ∗ A−1 X + Y ∗ B −1 Y ≥ (X + Y )∗ (A + B)−1 (X + Y ). Proof. By Lemma 1.4 we have   A X ≥ 0, X ∗ X ∗ A−1 X



B Y Y ∗ Y ∗ B −1 Y

(1.23)

 ≥ 0.

Applying the Schur product theorem gives   A◦B X ◦Y ≥ 0. (X ◦ Y )∗ (X ∗ A−1 X) ◦ (Y ∗ B −1 Y )

(1.24)

Applying Lemma 1.4 again in another direction to (1.24) yields (1.22). The inequality (1.23) is proved in a similar way.



Now let us consider some useful special cases of this theorem. Choosing A = B = I and X = Y = I in (1.22) respectively we get Corollary 1.20 For any X, Y and positive definite A, B (X ∗ X) ◦ (Y ∗ Y ) ≥ (X ◦ Y )∗ (X ◦ Y ),

(1.25)

A−1 ◦ B −1 ≥ (A ◦ B)−1 .

(1.26)

In (1.26) setting B = A−1 we get A ◦ A−1 ≥ (A ◦ A−1 )−1 or equivalently A ◦ A−1 ≥ I,

for A > 0.

(1.27)

(1.27) is a well-known inequality due to M. Fiedler. Note that both (1.22) and (1.23) can be extended to the case of arbitrarily finite number of matrices by the same proof. For instance we have k 

Xj∗ A−1 j Xj

≥

k 

1

Xj

∗ k 

1

Aj

−1 k 

1

Xj

1

for any Xj and Aj > 0, j = 1, . . . , k, two special cases of which are particularly interesting:

∗ k

k k    ∗ Xj Xj ≥ Xj Xj , k 1 k  1

A−1 j

1

≥k

2

k 

1

−1 Aj

,

each Aj > 0.

1

We record the following fact for later use. Compare it with Lemma 1.4.

1.4 Block Matrix Techniques



Lemma 1.21

A B B∗ C

15

 ≥0

(1.28)

if and only if A ≥ 0, C ≥ 0 and there exists a contraction W such that B = A1/2 W C 1/2 . Proof. Recall the fact that



I X X∗ I

 ≥0

if and only if X is a contraction. The “if” part is easily checked. Conversely suppose we have (1.28). First consider the case when A > 0, C > 0. Then   I A−1/2 BC −1/2 (A−1/2 BC −1/2 )∗ I    −1/2   −1/2 A B A 0 0 A ≥ 0. = B∗ C 0 C −1/2 0 C −1/2 Thus W ≡ A−1/2 BC −1/2 is a contraction and B = A1/2 W C 1/2 . Next for the general case we have   A + m−1 I B ≥0 C + m−1 I B∗ for any positive integer m. By what we have just proved, for each m there exists a contraction Wm such that B = (A + m−1 I)1/2 Wm (C + m−1 I)1/2 .

(1.29)

Since the space of matrices of a given order is finite-dimensional, the unit ball of any norm is compact (here we are using the spectral norm). It follows that there is a convergent subsequence of {Wm }∞ m=1 , say, limk→∞ Wmk = W. Of course W is a contraction. Taking the limit k → ∞ in (1.29) we obtain B = A1/2 W C 1/2 . 

Notes and References. Except Lemma 1.21, this section is taken from [86]. Note that Theorem 1.19 remains true for rectangular matrices X, Y. The inequality (1.22) is also proved independently in [84].